Answer
See below
Work Step by Step
The number of bacteria can be described by
$$P(t)=Ce^{kt}$$
with $C,k$ are constant.
Take the logarithm of each side:
$$\ln P(t)=kt+\ln C$$
Let $y=\ln P(t)\\
b=\ln C$
The data points are:
$(0,\ln 1.5);(12,\ln 5);(20,\ln 11);(24,\ln 16)$
Take the logarithm of number of bacteria $P(t)$:
$(0,0.405);(12,1.609);(20,2.398);(24,2.773)$
Form the matrices:
$A=\begin{bmatrix}
0 & 1\\
12 & 1\\
20 & 1\\
24 & 1\end{bmatrix}\\
x=\begin{bmatrix}
k\\b
\end{bmatrix}\\
b=\begin{bmatrix}
0.405\\
1.609\\
2.398\\
2.773
\end{bmatrix}$
Obtain:
$x_0=(A^TA)^{-1}A^Tb\\
=(\begin{bmatrix}
0 & 12 & 20 & 24 \\
1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
0 & 1\\
12 & 1\\
20 & 1\\
24 & 1\end{bmatrix})^{-1}\begin{bmatrix}
0 & 12 & 20 & 24\\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
0.405\\
1.609\\
2.398\\
2.773
\end{bmatrix}\\
=\begin{bmatrix}
1120& 56\\
56 & 4
\end{bmatrix}^{-1}\begin{bmatrix}
0 & 12 & 20 & 24 \\
1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
0.405\\
1.609\\
2.398\\
2.773
\end{bmatrix}\\
=\frac{1}{336}\begin{bmatrix}
1 & -14\\
-14 & 280
\end{bmatrix}\begin{bmatrix}
0 & 12 & 20 & 24 \\
1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
0.405\\
1.609\\
2.398\\
2.773
\end{bmatrix}\\
=\frac{1}{336}\begin{bmatrix}
-14 & -2 & 6 & 10 \\
280 & 112 & 0 & -56
\end{bmatrix}\\
=\frac{1}{336}\begin{bmatrix}
33.23\\
138.32\end{bmatrix}\\
=\begin{bmatrix}
0.099\\
0.412
\end{bmatrix}\\
k=0.099\\
b=0.412$
Since $b=\ln C\rightarrow C=e^b=e^{0.412}=1.51$
Consequently, $P(t)=Ce^{kt}=1.51e^{0.099t}$
